A quadratic function has x-intercepts at x=-6 and x=2. The point (-5,-14) is on the curve of the quadratic. What is the vertex of the quadratic?

Answer 1

To find the vertex of the quadratic function given the x-intercepts, we can use the fact that the vertex lies on the axis of symmetry of the parabola, which is halfway between the two x-intercepts.

Given the x-intercepts at $x=-6$ and $x=2$, we can calculate the x-coordinate of the vertex as follows:

$$\text{Vertex }x=\frac{-6+2}{2}=\frac{-4}{2}=-2$$

Next, we need to find the y-coordinate of the vertex. We can express the quadratic function in vertex form or standard form. Since we have the x-intercepts, we can initially write the quadratic in factored form as:

$$f(x)=a(x+6)(x-2)$$

We need to determine the value of $a$ using the additional point provided: (-5, -14).

Substituting $x=-5$ and $f(x)=-14$:

$$-14=a(-5+6)(-5-2)$$

This simplifies to:

$$-14=a(1)(-7)⟹-14=-7a⟹a=2$$

Now that we have $a$, the quadratic function becomes:

$$f(x)=2(x+6)(x-2)$$

Now we'll expand this to get it into standard form:

$$f(x)=2(x^2+4x-12)$$ $$=2x^2+8x-24$$

In standard form, the quadratic is $f(x)=2x^2+8x-24$.

To find the vertex, we can use the vertex formula for the x-coordinate:

$$x=-\frac{b}{2a}=-\frac{8}{2\cdot 2}=-\frac{8}{4}=-2$$

Now substituting $x=-2$ back into the quadratic function to find the y-coordinate of the vertex:

$$f(-2)=2(-2{)}^2+8(-2)-24$$ $$=2(4)-16-24$$ $$=8-16-24=-32$$

Thus, the vertex of the quadratic function is:

$$\boxed{{\displaystyle (-2,-32)}}$$